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blanchedsv
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Jack played a game of successive division in which he randomly chose a 13 Mar 2022, 13:51
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

53% (02:36) correct 47% (02:42) wrong based on 247 sessions

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Jack played a game of successive division in which he randomly chose a number and divided it by 2. He took the quotient from the last operation and divided it by 3. Then, he took the new quotient and divided it by 4. Similarly, he kept dividing the quotients with successive divisors until he got 0 as the quotient.

If the remainders in his game were 1, 1, 3, 2, 4 (in that order), which of the following was the initial number that he had randomly chosen?

A. 69
B. 549
C. 739
D. 1269
E. 2589
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B
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mbencan
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Jack played a game of successive division in which he randomly chose a 13 Mar 2022, 19:01
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Ans. B

If q=0 with remainder of 4 - when x:6 - x=4
If q=4 with remainder of 2 - when x:5 - x=22
If q=22 with remainder of 3 - when x:4 - x=91
If q=91 with remainder of 1 - when x:3 - x=274
If q=274 with remainder of 1 - when x:2 - 549

You can also check the result as follows:

549:2 = 274 + 1
274:3 = 91 + 1
91:4 = 22 + 3
22:5 = 4 + 2
4 : 6 = 0 + 4
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Jack played a game of successive division in which he randomly chose a 18 Mar 2022, 02:37
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Given: Jack played a game of successive division in which he randomly chose a number and divided it by 2. He took the quotient from the last operation and divided it by 3. Then, he took the new quotient and divided it by 4. Similarly, he kept dividing the quotients with successive divisors until he got 0 as the quotient.

Asked: If the remainders in his game were 1, 1, 3, 2, 4 (in that order), which of the following was the initial number that he had randomly chosen?

A. 69: 69 = 2*34+1; 34 = 3*11+1; 11= 2*4+3; 4 = 0*5 + 4; Incorrect
B. 549; 549 = 2*274 + 1; 274 = 3*91 + 1; 91 = 4*22 + 3; 22 = 5*4 + 2; 4 = 0*5 + 4; Correct
C. 739
D. 1269
E. 2589

No need to test the remaining answer choices

IMO B
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Jack played a game of successive division in which he randomly chose a 17 Jun 2023, 09:48
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blanchedsv
Jack played a game of successive division in which he randomly chose a number and divided it by 2. He took the quotient from the last operation and divided it by 3. Then, he took the new quotient and divided it by 4. Similarly, he kept dividing the quotients with successive divisors until he got 0 as the quotient.

If the remainders in his game were 1, 1, 3, 2, 4 (in that order), which of the following was the initial number that he had randomly chosen?

Dividend = Divisor * Quotient + Remainder
D = x * Q + R

According to question,
R 4 2 3 1 1
x 6 5 4 3 2
Q 0
D 4

Now, Quotient = Dividend
and calculating for dividend and repeating the process, we get the following matrix.

So,
R--4---2---3----1----1
x--6---5---4----3----2
Q--0---4--22---91--274
D--4--22--91--274--549

Hence, Option (B)
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Jack played a game of successive division in which he randomly chose a 12 Jul 2023, 03:44
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Kinshook
Given: Jack played a game of successive division in which he randomly chose a number and divided it by 2. He took the quotient from the last operation and divided it by 3. Then, he took the new quotient and divided it by 4. Similarly, he kept dividing the quotients with successive divisors until he got 0 as the quotient.

Asked: If the remainders in his game were 1, 1, 3, 2, 4 (in that order), which of the following was the initial number that he had randomly chosen?

A. 69: 69 = 2*34+1; 34 = 3*11+1; 11= 2*4+3; 4 = 0*5 + 4; Incorrect
B. 549; 549 = 2*274 + 1; 274 = 3*91 + 1; 91 = 4*22 + 3; 22 = 5*4 + 2; 4 = 0*5 + 4; Correct
C. 739
D. 1269
E. 2589

No need to test the remaining answer choices

IMO B

HI Kinshook

in similar fashion as tried by you

1269= 2*634+1 ,
634=3*211+1
211=4*52+3
52=5*10+2
10=6*1+4
4=7*0+4


Then why option D is not the correct answer too ??

Thx
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Jack played a game of successive division in which he randomly chose a 14 Jul 2023, 01:31
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blanchedsv
Jack played a game of successive division in which he randomly chose a number and divided it by 2. He took the quotient from the last operation and divided it by 3. Then, he took the new quotient and divided it by 4. Similarly, he kept dividing the quotients with successive divisors until he got 0 as the quotient.

If the remainders in his game were 1, 1, 3, 2, 4 (in that order), which of the following was the initial number that he had randomly chosen?

A. 69
B. 549
C. 739
D. 1269
E. 2589

Solution:

  • Let the chosen number be \(x\)
  • Now according to the question:
    \(x=2Q_1+1\)
    \(Q_1=3Q_2+1\)
    \(Q_2=4Q_3+3\)
    \(Q_3=5Q_4+2\)
    \(Q_4=6Q_5+4\)
  • This division stops when the quotient becomes 0 which means \(Q_5=0\). Thus,
    \(Q_4=6Q_5+4=6\times 0+4=4\)
    \(Q_3=5Q_4+2=5\times 4+2=22\)
    \(Q_2=4Q_3+3=4\times 22+3=91\)
    \(Q_1=3Q_2+1=3\times 91+1=274\)
    \(x=2Q_1+1=2\times 274+1=549\)

Hence the right answer is Option B
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Jack played a game of successive division in which he randomly chose a 15 Jun 2025, 12:08
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An interesting problem statement, where the catch is to solve the whole question backwards ie.

Count the number of remainders => 1 1 3 2 4 and remember their relevant divisors => 2 3 4 5 6

Now we know that Jack stopped playing when he got to quotient 0 which means,

Last dividend was 6*0 + 4 = 4

We know second last divisor = 5, second last remainder = 2 and second last quotient = 4. This helps us calculate second last dividend,

Second last divided - 5*4 + 2 = 22

We will repeat this till we get the initial number =>

Third last dividend - 4*22 + 3 = 91
Fourth last dividend - 3*91 + 1 = 274
Initial number - 2*274 + 1 = 549

IMO: B
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